In a List<Object>, how can I keep the most recent entry and remove previous entries which have an identical field?

Question

I have a List which contains a record of various CallLog objects. Every CallLog is different, and when I pass this to my ListView, each log is correctly displayed on my screen. What I want to do, however, is remove logs which have the same RemoteAddress attribute (ie a person who called me or has received a call by me should only appear once in my ListView). This is because I will display all log details for that remote contact directly below it.

How can I create a method (or class), which could filter out my List to only keep unique Remote Addresses? Below is how I retrieve this log list from the core. It has the form of a CallLog[]:

// Filter this
List<CallLog> mLogs = Arrays.asList(LinphoneManager.getCore().getCallLogs());

Answer 1

List<CallLog> mLogs = Stream.<CallLog>of(LinphoneManager.getCore().getCallLogs())
            .collect(ArrayList::new, 
                (a, l) -> { if (!a.stream().anyMatch(o -> ((CallLog) o).remoteaddress.equals(l.remoteaddress))) a.add(l); }, 
                (a, b) -> { a.addAll(b); })

Have you tried something like this before? You can use set method rather than add.